I have a homework problem in which I wish to show that the family of curves given by

$x^2 + y^2 = c x$

where $c$ is an abitrary constant may be described by the differential equation

$\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$

I thought that I could use implicit differentiation to differentiate the original equation to get the second equation, but instead I get the equation

$\frac{c-2x}{2y}=\frac{dy}{dx}$

As you can see the derivative I get is not in the form of the equation that I am supposed to get. I do not see a way for my solution to even become similar to the proposed solution as one contains constants whereas the other does not.

4 Answers
4

This is what I got too: differentiate $x^2 + y^2 = c x$ with respect to $x$, we get
$$2x + 2y\frac{dy}{dx} = c, $$
which implies that
$$\frac{dy}{dx}=\frac{c-2x}{2y},$$
as you got. But you can multiplying $x$ to get:
$$\frac{dy}{dx}=\frac{c-2x}{2y}\cdot\frac{x}{x}=\frac{cx-2x^2}{2xy}=\frac{(x^2 + y^2 )-2x^2}{2xy}= \frac{y^2-x^2}{2xy},$$
where we have used $cx=x^2 + y^2$ in the third equality.

On the other hand, you can also solve the differential equation: $\displaystyle\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$. Let $y=tx$. Then $dy=tdx+xdt$. Therefore the differetial equation can be written as
$$tdx+xdt=\frac{t^2x^2-x^2}{2tx^2}dx=\frac{t^2-1}{2t}dx,$$
which implies that
$$xdt=-\frac{t^2+1}{2t}dx.$$
Integrating it, we get
$$\ln(t^2+1)=\int\frac{2t}{t^2+1}dt=-\int\frac{dx}{x}=-\ln x+C.$$
Hence, we have
$\ln[x(t^2+1)]=C$. Recall $t=y/x$, we get
$\displaystyle\frac{y^2}{x}+x=e^C.$ If we write $e^C=c$, we get
$$y^2+x^2=cx.$$

First multiply by $\frac{x}{x}$ $$\frac{c-2x}{2y}\cdot\frac{x}{x}=\frac{x(c-2x)}{2xy}=\frac{cx-2x^2}{2xy}$$ From the first equation we know that $$cx=x^2+y^2$$ so $$\frac{cx-2x^2}{2xy}=\frac{(x^2+y^2)-2x^2}{2xy}=\frac{y^2-x^2}{2xy}$$ Therefore $$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$